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Anonymous users2024-02-07Proof: Because (a-b) 2>0, (b-c) 2>0, (c-a) 2>0, (a-b) 2+(b-c) 2+(c-a) 2>0a 2-2ab+b 2+b 2-2bc+c 2+c 2-2ac+a 2>0
2(a^2+b^2+c^2-ab-bc-ca)>0a^2+b^2+c^2>ab+bc+ca
Therefore, the square of a + the square of b + the square of c ab + bc + ca The formula of the second question is not understood, please make it clearer.
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Anonymous users2024-02-06The square of a + the square of b 2ab>0
The square of b + the square of c 2bc>0
The square of a + the square of c is 2ac>0
The sum of the three forms is proven.
Question 2 is ab+cd, right?
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Anonymous users2024-02-05A 2 + VB 2 + C 2 + D 2 A 2 * B 2 * C 2 * D 2 i.e.
a^2+vb ^2+c^2+d^2≥？
It's hard, I'm sorry, I don't know how to do it.
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Anonymous users2024-02-04c=5, Youchun.
b/a=1/2
c²=a²+b²
25=5b²
b²=5a²=20
Equation x 20-y 5 = 1
y=kx+b double good service curve.
And y1 talk about (x1-2)*y2 (x2-2)=-13b 2+16kb+36k 2+16=0
The discriminant formula is constant 0, and only k=0 is problematic?
The idea is to find the relationship between k and b and find the fixed point.
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Anonymous users2024-02-03The answer is correct.
The process is cumbersome, you must learn to calculate lazily, otherwise you will die of exhaustion as follows: 1+k 2=5
x1+x2=-12m/10,x1x2=(3m^2+6)/10(x1-x2)^2=6(m^2-10)/25ab^2=(x1-x2)^2(1+k^2)16=6(m^2-10)*5/25
m^2=70/3
m=±√210/3
ab;y=2x±√210/3
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Anonymous users2024-02-021 All modes is the middle of the highest rectangular corresponding to the abscissa.
That is, the median is the value corresponding to half of the probabilities, and the probability of the first rectangle is .
To make up half of the probability, it is needed.
The probability of the second rectangle corresponds to is equal to.
The rest is inside the second rectangle.
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Anonymous users2024-02-01The number on the vertical axis is the frequency divided by the group spacing.
f（5~10）=
f（10~15）=
f（15~20）=
The mode is the highest frequency of 10 15, take the average, for.
The median is to be sorted and falls in the second half of the group of 10 to 15, which is equal to 10+ (15-10).
So, the mode is, and the median is 13
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Anonymous users2024-01-31The middle of the interval with the most modes [10-15 is.
The median number is the area.
And the area of the frequency distribution histogram is one.
Therefore, the corresponding number when the area is taken is the median.
x = 13 median is thirteen.
It's a bit hard to type qaq
This can be solved using inequalities.
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